Why List partition works while span does not

Question

I have this List oX of (Char, Int) pars (it contains pairs with only unique Char values)

List(( ,3), (d,1), (e,3), (h,3), (i,1) , (l,3), (o,2), (r,2), (t,1), (w,1))

I need to partition this list into 2 - one that contains any pair that has 'd' Char and another is the rest.

So I tried partition and span but found that span does not work as expected. Here are the results (copied from scala worksheet)

 val myPartition = oX.partition(e => e._1 == 'd') > myPartition  : (List[(Char, Int)], List[(Char, Int)]) = (List((d,1)),List(( ,3), (e,3), (h,3), (i,1), (l,3), (o,2), (r,2), (t,1), (w,1)))

  val mySpan = oX.span(e => e._1 == 'd') > mySpan  : (List[(Char, Int)], List[(Char, Int)]) = (List(),List(( ,3), (d,1), (e,3), (h,3), (i,1), (l,3), (o,2), (r,2), (t,1), (w,1)))

I am puzzled why given same predicate functino partition gives me expected result while span gives me empty List as first list and copy of original as second list

Answer 1

From the documentation of span: "Returns the longest prefix of the list whose elements all satisfy the given predicate, and the rest of the list."

So here span gives you the expected result: since the first tuple in your list doesn't have the character d, the longest prefix of the list where each tuple has the character d is indeed the empty list.